### Q.17:- A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

**Answer:-**

Length of the string, *l* = 80 cm = 0.8 m

Number of revolutions = 14

Time taken = 25 s

Frequency, *v* = Number of revolutions / Time taken = 14 / 25 Hz

Angular frequency, ω = 2πν

= 2 × (22/7) × (14/25) = 88 / 25 rad s^{-1}

Centripetal acceleration, *a _{c}* = ω

^{2}

*r*

= (88/25)

^{2}× 0.8

= 9.91 ms

^{-2}

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.